The graph shows the solution to the initial value problem

The graph shows the solution to the initial value problem , . Find the following.

Solution

dy/dt = mt

dy = mtdt

Integrating :

y = mt^2/2 + C

From the given graph, it is clearly a parabola with vertex at (0 , -2)

So, it now looks like :

y = a(t - 0)^2 - 2

y = at^2 - 2

Also, there is a point (2,1) :

1 = a(2)^2 - 2

1 = 4a - 2

3 = 4a

a = 3/4

So, the solution is :

y = (3/4)t^2 - 2

Comparing with y = mt^2/2 + C, we get :

m/2 = 3/4

m = 3/2 --> ANSWER

C = -2

y(t0) = -2 :

(3/4)(t0)^2 - 2 = -2

(3/4)(t0)^2 = 0

t0^2 = 0

t0 = 0

So, answers are :

m = 3/2
to = 0
y(t) = (3/4)t^2 - 2